3.357 \(\int (b \sec (e+f x))^m \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=63 \[ \frac{\tan ^5(e+f x) \cos ^2(e+f x)^{\frac{m+5}{2}} (b \sec (e+f x))^m \, _2F_1\left (\frac{5}{2},\frac{m+5}{2};\frac{7}{2};\sin ^2(e+f x)\right )}{5 f} \]

[Out]

((Cos[e + f*x]^2)^((5 + m)/2)*Hypergeometric2F1[5/2, (5 + m)/2, 7/2, Sin[e + f*x]^2]*(b*Sec[e + f*x])^m*Tan[e
+ f*x]^5)/(5*f)

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Rubi [A]  time = 0.0395067, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {2617} \[ \frac{\tan ^5(e+f x) \cos ^2(e+f x)^{\frac{m+5}{2}} (b \sec (e+f x))^m \, _2F_1\left (\frac{5}{2},\frac{m+5}{2};\frac{7}{2};\sin ^2(e+f x)\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[(b*Sec[e + f*x])^m*Tan[e + f*x]^4,x]

[Out]

((Cos[e + f*x]^2)^((5 + m)/2)*Hypergeometric2F1[5/2, (5 + m)/2, 7/2, Sin[e + f*x]^2]*(b*Sec[e + f*x])^m*Tan[e
+ f*x]^5)/(5*f)

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

Rubi steps

\begin{align*} \int (b \sec (e+f x))^m \tan ^4(e+f x) \, dx &=\frac{\cos ^2(e+f x)^{\frac{5+m}{2}} \, _2F_1\left (\frac{5}{2},\frac{5+m}{2};\frac{7}{2};\sin ^2(e+f x)\right ) (b \sec (e+f x))^m \tan ^5(e+f x)}{5 f}\\ \end{align*}

Mathematica [C]  time = 29.413, size = 10908, normalized size = 173.14 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*Sec[e + f*x])^m*Tan[e + f*x]^4,x]

[Out]

Result too large to show

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Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int \left ( b\sec \left ( fx+e \right ) \right ) ^{m} \left ( \tan \left ( fx+e \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^m*tan(f*x+e)^4,x)

[Out]

int((b*sec(f*x+e))^m*tan(f*x+e)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e))^m*tan(f*x + e)^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec{\left (e + f x \right )}\right )^{m} \tan ^{4}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))**m*tan(f*x+e)**4,x)

[Out]

Integral((b*sec(e + f*x))**m*tan(e + f*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \sec \left (f x + e\right )\right )^{m} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(f*x+e))^m*tan(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e))^m*tan(f*x + e)^4, x)